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Integral of Arcsin[x]

  1. Sep 15, 2005 #1
    I just can't see it. I would think that it would involve some form of trig substitution, but I'm just drawing a blank. I'll do the work if someone can please give me a nice little hint.

    I know that [tex]\int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C[/tex] from my calculator and mathematica.

    Hint please. :yuck:
  2. jcsd
  3. Sep 15, 2005 #2


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    If only you were working with its derivative! I bet you know how to integrate something that looks like its derivative!
  4. Sep 15, 2005 #3
    Sure... I'd draw a nice triangle and do trig substitution if it was [tex]\int \frac{dx}{\sqrt{1-x^2}}[/tex]

    Hmmm...I'll think out loud here. If the integral has a square-root and is in the form of [itex]c^2-x^2[/itex] then x is one leg, c is the hypotenuse, and the other leg is the previously mentioned radical.

    Is trig substitution with right triangles on the right track? Since the hypotenuse is [itex]\sqrt{a^2+b^2}[/itex], it seems that one leg might need to be [itex]\sqrt{\arcsin{(x)}}[/itex] Somehow I don't think I'm on track.
  5. Sep 15, 2005 #4
    Use integration by parts. Remember to let u=arcsin(x) and v=x.

    u sub: InverseLogAlgebraicTrigExp
  6. Sep 15, 2005 #5


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    Sheesh, just give him the answer, why don't ya? :grumpy:
  7. Sep 15, 2005 #6
    Ah, thank you! :surprised It's so simple.

    And thank you Hurkyl as well. I still have lots to learn.
  8. Sep 16, 2005 #7
    I apologize Hurkyl :smile:
  9. Sep 16, 2005 #8
    well, you know the integral of sinx with limits. Now arcsin x will be the limits, and you can make a rectangle.
  10. Sep 16, 2005 #9
    Or you could just take the derivative of the right hand side and go "ta da!" and that's proof enough for me.
  11. Sep 17, 2005 #10

    equate the arc sine to another variable e.g y.making it a sine fxn.e.g the arc sine of 0.5=30,while sine30=0.5.this will simplify the integral and further substitution will conclude it
  12. Sep 19, 2005 #11


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    Just for the fun of it ...

    The sum of the integrals

    [tex]\int \sin^{-1} x dx + \int \sin y dy[/tex]

    is just the area of the bounding rectangle: [itex] x \times \sin^{-1} x[/itex]

    Since [tex]\int \sin y dy = -\cos y + C[/tex] and [tex]\cos y = \cos \sin^{-1} x = \sqrt {1-x^2}[/tex] it follows that
    [tex]\int \sin^{-1} x dx = \sqrt {1-x^2} + x \sin^{-1} x + C[/tex]
  13. Jun 9, 2007 #12
    Integration by parts?

    Hey, Im really sorry to arise dead threads from the past (which i have seen though google) but somehting really weird happened me when I tried to use integration by parts on arcsinx.

    let me show you:
    S(arcsinx)= {v'(x)=1} {u(x)=arcsinx}


    xarcsinx-S(x/(1-x^2)^0.5= {u(x)=x v'(x)=arcsinx}


    ==> S(arcsinx)=S(arcsinx)

    I know I have done something really stupid here, but please be easy on me since I started studying Integrals only three days ago.

    In the second time I used integration by parts, do I miss something , is there another efficient choice of v and u?

    Thanks in advance,

    p.s: I will edit it better to use normal math signs once I figure out how.
  14. Jun 9, 2007 #13

    Gib Z

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    For the second time you integrate by parts, swap your choices.
  15. Jun 9, 2007 #14
    Im sorry, Tried it also and all i got is:

    xarcsine x -(x^2/2)(1/(Sqrt(1-x^2)))-(x/4)(sqrt(1/(1-x^2))+1/4(arcsinx)

    this isn't going anywhere :(
  16. Jun 9, 2007 #15


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    Make [itex] x=\sin t [/itex]. Then apply part integration on the resulting integral. It's just a way to avoid the simple solution of part integrating directly.
  17. Jun 9, 2007 #16


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    After the first integration by parts, I would use a substitution
  18. Jun 9, 2007 #17
    solve it officially. Fixed during some major mistakes I had about dev' and stuff.

    did it without subtition, only using integration by parts.

    if you are interested what I did then you are welcome to tell me to write my solution.

    Thanks guys :)
  19. Oct 10, 2008 #18
    use seperation by parts



    uv-ingegral vdu = xarsinx-integral x/(1-x^2)^1/2
    use u substitution with u = 1-x^2 so du = -2x
    than you get

    xarsinx + (1-x^2)^1/2 + c
  20. Apr 27, 2010 #19
    f(x)=arcsinx f'(x)=1/radical(1-xsquare)
    g'(x)=1 g(x)=x

    // S means integral

    S arcsin x dx= x arcsin x - S xdx/radical(1-xsquare) = x arcsin x + S (radical(1-xsquare))'dx=
    =x arcsin x + radical(1-xsquare) +C
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